In a hydraulic garage lift, the small piston has a radius of 5.0 cm and the large piston has a radius of 1.50 m. what force must be to the small piston in order to lift a zamboni weighing 200,000 n on the large piston?

222.22 N

Explanation:

r = 5 cm = 0.05 m, R = 1.5 m, f = ? , F = 200,000 N

Use pascal's law

F / A = f / a

F / 3.14 x R^2 = f / 3.14 x r^2

F / R^2 = f / r^2

f = F x r^2 / R^2

f = 200000 x 0.05 x 0.05 / (1.5 x 1.5)

f = 222.22 N

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

massxgravity x heght=weightxheight= pe

4900=490xh

h=10